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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: pear on November 19, 2009, 12:04:45 AM

Title: electrochemistry; H3O+ + OH- ---> 2H2O
Post by: pear on November 19, 2009, 12:04:45 AM

Hello!

I am having difficulty understanding the following problem.
    Write cathode & anode 1/2-reactions for the following Bronsted neutralization reaction:
    H₃O⁺(aq) + OH^-  :rarrow: 2H₂O(l)


What I wrote before checking the solution manual:
    H₃O⁺  :rarrow:  H₂O(l) + H⁺
    OH^-  + H⁺  :rarrow:  H₂O(l)
but quickly recognized a lack of e^- exchange.


Well, the solution manual says:
    rewrite as:  H⁺(aq) + OH^-(aq)  :rarrow:  H₂O(l)
    oxidation@anode:  O₂(g) + 2H₂O(l) + 4e^-  :rarrow:  4OH^-(aq)
    reduction@cathode:  O₂(g) + 4H⁺ + 4e^-  :rarrow:  2H₂O(l)

So.. How do I know to do the re-write?
Why does ...
    H⁺  :rarrow:  H₂O(l)
not follow the procedure balance: A) elements other than H & O;  B) O w/ H₂O;  C) H w/ H⁺;  D) charge w/ e, 
which would give you (at the anode) ...
    OH^- + H⁺  :rarrow:  H₂O  ?


Thanks in advance!

Title: Re: electrochemistry; H3O+ + OH- ---> 2H2O
Post by: Borek on November 19, 2009, 03:37:36 AM

I don't like the question.

Quote from: pear on November 19, 2009, 12:04:45 AM

but quickly recognized a lack of e^- exchange.

Hardly suprprising - there is NO electron exchange, as this is not a redox reaction.

I guess they wan't to separate reactants yet somehow allow for the neutralization to occur. For that you need two separate reactions - one consuming H⁺, other one consuming OH^-. Net reaction will be neutralization, but for me that's putting things on the head.